Craps - Multiple Wagers House Edge
Poll
 | 2 votes (33.33%) | ||
| | 1 vote (16.66%) | ||
| | 2 votes (33.33%) | ||
| | 1 vote (16.66%) |
6 members have voted
From a famous post here:
https://wizardofvegas.com/forum/gambling/craps/11091-best-bets-with-small-bank-roll/2/#post175132
I want to use these two.
#1 question/answer shown below is 1.33% house edge
for a $25 Buy on the #4 and paying a $1 commission on a win and taking the bet down after the win.
#2 q/a shown below is 1.00% house edge
for a $25 Buy on the #4 and a $25 Buy on the #10,
paying a $1 commission on a win and taking both bets down after the win.
Only one bets wins but both bets could lose.
Quote: Ahigh
#1) Buy the 4 for $25 take $74 and down 1/75 is 1.33% ...
#2) But [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
There is no prize for correct answers.
A good formula to use would be EV / Action = house edge
But one can use any formula.
For those willing to try but do not know the payoffs and probabilities... here they be
Buy4: prob winning 1/3
prob of a loss 2/3
Payout 2 to 1 on a win. (So, $25 pays $50)
Commission charged will be $1. This means only $49 is the total actual win.
Vote, state your answers and Show your work.
You do not have to state your answers or even show your work.
That will come after many have had the chance to Vote.
No one is wrong and everyone will be right by stating their answers.
This is for all those that really want to know how to do this properly.
The internet is filled with lots of mis-information.
Let the Wizard of Vegas forum stand out and show how this is done for a Craps bet.
Some can then take that knowledge, if wanted, and apply it to other bets outside of Craps.
The importance I see for knowing the proper way to calculate the house edge is we need HE to determine the EV, and with the variance one can get a good picture of ruin and win goal probabilities from any length of play from it.
Improper house edge can underestimate the actual ev and the more trials, the greater the error becomes.
Enjoy
Quote:#1) Buy the 4 for $25 take $74 and down 1/75 is 1.33% ...
3 ways to roll the FOUR, 6 ways to roll the SEVEN = 1:2
(.333*49 - .666*25)/(25) = 1.33% HE
(ways to win x amount won - ways to lose x amount loss) / (Amount of Action)
On a roll of 7, lose the $25, on a roll of 4 take the $49 win + take down the $25 bet = $74
Quote:#2) Buy [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
3 ways to roll the FOUR and 3 ways to roll the 10 = 6 ways to win, 6 ways to roll the SEVEN = 1:1
(.5*49 - .5*50)/(50) = 1.00% HE
(ways to win x amount won - ways to lose x amount loss) / (Amount of Action)
On a roll of 7, lose $50 , on a roll of 4 or 10, one bet wins $49, the other bet is NO ACTION, and both bets ($50 total) are taken down = $99.
Because the other bet had NO ACTION, it should not be included in the Action denominator.
-------------------------------------------------------------------------------------------------------------------------------
It all comes down to, is there a house edge on a bet that was made, but was cancelled prior to a resolution?
So in scenario #2, I have $50 at risk, with a 50-50 chance of winning $49, and a 50-50 chance of losing $50.
If I have the bet active for 4 rolls with no resolution, and then pull it down, was the bet subject to a HE ?
What I could not resolve from my post above was the following;
The Buy 4 by itself is 1.33%, and the Buy 10 by itself is 1.33%.
So how can I combine those 2 bets and have a HE less than the 1.33%? That makes no sense.
3 ways to roll the FOUR and 3 ways to roll the 10 = 6 ways to win, 6 ways to roll the SEVEN = 1:1
Quote:(.5*49 - .5*50)/(50) = 1.00% HE
(ways to win x amount won - ways to lose x amount loss) / (Amount of Action)
This is not right. The $50 in Action from the above equation does not account for the fact that I have $25 risked against the HE for the time while the bet is active. So, even though I pull it down, it was at risk for part of the time.
Therefore, the total Action should be:
50% of the time (a 7), I will lose $50.
25% of the time (a 4), I will win on $25 of Action, and take $25 off the table
25% of the time (a 10) I will win on $25 of Action and take $25 off the table.
written as (.25*$25+.25*$25+.50*$50)
so the correct equation reads
(.5*49- .5*50)/(.25*$25+.25*$25+.50*$50) = 1.33%
------------------------------------------------------------------------------------------------
For consistency, shouldn't the single Buy 4 bet probably be written like this;
(.333*$49 - .666*$25) / (.333*$25+.666*$25) = 1.33%
1/3 of the time I win $49
2/3 of the time I lose $50
1/3 of the time when I win I have $25 in Action.
2/3 of the time when I lose I have $25 in Action.
That's because you did not factor in Dice Setting !
Quote: buzzpaff" So how can I combine those 2 bets and have a HE less than the 1.33%? That makes no sense."
That's because you did not factor in Dice Setting !
I didn't see Dice Setting listed in the OP's original question. How would I add that into the equation?
Is this question _really_ so hard that you have to resort to a poll?
If you think you have two bets you are wrong.
If you have two bets, explain to me how you win both of them? Roll 4 dice?
You guys are hilarious.
Why not just get religious and do a holy prayer that you are right and I am wrong.
The fact is, and you will finally learn eventually, is that I am right and you in fact are wrong.
Just move along and admit you are wrong.
It's been weeks now and you _still_ haven't figured out that I'm right?
This is HILARIOUS.
Yeah, and keep on using dice influence as part of your argument that my math is wrong.
That's a really strong argument there.
Seriously!!!
ROFMFL.
Quote: buzzpaffAt least dice setters know when they are lying !
Just admit you're wrong.
Search for mensa, and read all that and get back to me.
https://wizardofodds.com/ask-the-wizard/craps/betting-systems/
Kaythxbye!
AND .. it's in the eff aye queue. OMFG. Seriously. How many weeks now?
Initially I thought you were right with the 1.00% HE, but I had a couple of things that were bothering me, so I logically worked through the bet.
Please see my second post above, and I am genuinely asking you to tell me where my logic has gone wrong. I am serious here. I truly want to learn this stuff. I am not a math guy and I have no preformed or learned opinions. As I worked through it logically, I came to the conclusion that while we stand to lose $50 on a 7, on the other 2 numbers (4 and 10), only $25 is really in play, therefore our real Action is something less than the $50 that we can lose when the 7 is rolled.
Again, I am asking you to please quote my 2nd post and then show me where my logic has gone wrong, so that I may continue to learn. Thanks
The question #2 is difficult.Quote: AhighA poll????
Is this question _really_ so hard that you have to resort to a poll?
If you think you have two bets you are wrong.
If you have two bets, explain to me how you win both of them? Roll 4 dice?
You already gave your answer and there are two that do not agree with you, one is me.
(I had to do a simulation just to see, and I saw)
There are two separate bets made. Here is a pic of what it looks like.
Both can NOT win on the same roll, absolutely impossible, but both CAN LOSE on the same roll.
My wife IS a Mensa member, and she can not solve it and this has never been solved AFAIK by Mensa, it is that difficult.
RCraps, do you agree with ME on his HE calculation for Buy4and10 for $5000 each, that it is 1.67%?
https://wizardofvegas.com/forum/gambling/betting-systems/11356-what-are-the-odds/3/#post183638
His formula has me scratching my head. It is almost like yours but he puts 0.5*9750 in place of 0.5*4750
extra credit there
The Mensa people calculation is house edge per roll, not per resolution.Quote: AhighJust admit you're wrong.
Search for mensa, and read all that and get back to me.
https://wizardofodds.com/ask-the-wizard/craps/betting-systems/
Kaythxbye!
That is because some bets resolve on one roll and others on multiple rolls. They should have also calculated it per resolution to get the actual simulated value.
With 2 Buy bets, they both can resolve on a loss together but not on a win.
they will never be resolved on every roll.
still time
Quote: guido111A Craps house edge calculation.
From a famous post here:
https://wizardofvegas.com/forum/gambling/craps/11091-best-bets-with-small-bank-roll/2/#post175132
I want to use these two.
#1 question/answer shown below is 1.33% house edge
for a $25 Buy on the #4 and paying a $1 commission on a win and taking the bet down after the win.
#2 q/a shown below is 1.00% house edge
for a $25 Buy on the #4 and a $25 Buy on the #10,
paying a $1 commission on a win and taking both bets down after the win.
Only one bets wins but both bets could lose.Quote: Ahigh
#1) Buy the 4 for $25 take $74 and down 1/75 is 1.33% ...
#2) But [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
There is no prize for correct answers.
A good formula to use would be EV / Action = house edge
But one can use any formula.
For those willing to try but do not know the payoffs and probabilities... here they be
Buy4: prob winning 1/3
prob of a loss 2/3
Payout 2 to 1 on a win. (So, $25 pays $50)
Commission charged will be $1. This means only $49 is the total actual win.
Vote, state your answers and Show your work.
You do not have to state your answers or even show your work.
That will come after many have had the chance to Vote.
No one is wrong and everyone will be right by stating their answers.
This is for all those that really want to know how to do this properly.
The internet is filled with lots of mis-information.
Let the Wizard of Vegas forum stand out and show how this is done for a Craps bet.
Some can then take that knowledge, if wanted, and apply it to other bets outside of Craps.
The importance I see for knowing the proper way to calculate the house edge is we need HE to determine the EV, and with the variance one can get a good picture of ruin and win goal probabilities from any length of play from it.
Improper house edge can underestimate the actual ev and the more trials, the greater the error becomes.
Enjoy
Just to have a record of the original post... LOLz.
You have to be a math guy, you walked it thru.Quote: RaleighCraps
Please see my second post above, and I am genuinely asking you to tell me where my logic has gone wrong. I am serious here. I truly want to learn this stuff. I am not a math guy and I have no preformed or learned opinions. As I worked through it logically, I came to the conclusion that while we stand to lose $50 on a 7, on the other 2 numbers (4 and 10), only $25 is really in play, therefore our real Action is something less than the $50 that we can lose when the 7 is rolled.
Again, I am asking you to please quote my 2nd post and then show me where my logic has gone wrong, so that I may continue to learn. Thanks
Sometimes I am a math guy, but not always,
I cheat and do a simulation to see what my answer should be, and sometimes I even code that wrong.
RaleighCraps, now calculate the HE for the 2 Buy bets at $5000 each and see what you get.
I have to walk the dog
The edge per roll is the same, OBVIOUSLY.
But when the average number of rolls diminish, the edge per event also diminishes.
I can't help but find humor in how long this is taking WITH ME LEADING YOU AND POINTING TO FAQ's.
But please, continue to pretend that I am wrong if it helps you get through the emotional parts of being wrong.
Edge per roll is 0.33333 for both bets.
Average number of rolls is 4 for just one numbers.
Average number of rolls is three when either number causes the bet to be over.
.333333 * 3 = 1.00
.333333 * 4 = 1.33
OMFG numbers!
But that's EXACTLY what all of this is. But I do feel sorry for guys who's identity is to be math guys but can't do the most simple math.
Perhaps in your postal exchange with others, you missed my plea for help with my answer in the third post of this thread. I do not consider it a MATH post. I looked at it logically, and using logic, I have come up with an answer that does not agree with yours. Would you be so kind as to point out where I went wrong?
I would hate to be mislead by the majority here. I am already in enough trouble for my insistence on Buy bets in lieu of Come bets as it is....
You did great.Quote: RaleighCrapsAhigh,
Perhaps in your postal exchange with others, you missed my plea for help with my answer in the third post of this thread. I do not consider it a MATH post. I looked at it logically, and using logic, I have come up with an answer that does not agree with yours. Would you be so kind as to point out where I went wrong?
I would hate to be mislead by the majority here. I am already in enough trouble for my insistence on Buy bets in lieu of Come bets as it is....
Your answer is per bet resolved. I see is as correct.
AHigh now shows two answers, resolved and per roll. He is correct on both.
1000 bets * $bet * HE = EV
Most use the 1.33% for the HE in the EV formula.
A bit more math involved trying to use the 1% HE value that is per roll when one makes 1000 such bets.
Or very wrong values if one does use the 1% in this case.
Quote: guido111You have to be a math guy, you walked it thru.
Sometimes I am a math guy, but not always,
I cheat and do a simulation to see what my answer should be, and sometimes I even code that wrong.
RaleighCraps, now calculate the HE for the 2 Buy bets at $5000 each and see what you get.
I have to walk the dog
I am assuming that the formula that I took from another posting is, in fact, THE correct formula.
(expected win - expected loss)/ Action = HE
From there, I used logic to determine what I felt the correct numbers should be for each variable.
Are there other acceptable ways to calculate HE?, OR
more importantly, is there a better way to calculate Expected Outcome?, since that is really all I care about.
What is the Expected Outcome if I Buy the 4 and 10 for $50 each, and take them both down on a Win of either number?
----------------------------------------------------------------------------------------------------------------------
same formula, different numbers......
$5000 Buy 4, pays $10,000 - $250 Vig = $9750
(.25*9750 + .25*9750 - .5*10000) / (.25*5000 + .25*5000 + .5*10000) = 1.667% HE
There is no abuse here.Quote: AhighTrust and know that I never expect an apology for all this abuse.
But that's EXACTLY what all of this is.
But I do feel sorry for guys who's identity is to be math guys but can't do the most simple math.
I asked an honest question based on your original post and answers.
edit: The 1% still does not fit anywhere per bet or per roll.
You showed three answers and two others showed different answers.
Each thought both were wrong, but turns out both were right.
But you still will say only you are right.
Fine, Rock and Roll!
Solution, bet resolved and per roll
I stated at the bottom of my first post
"The importance I see for knowing the proper way to calculate the house edge is we need HE to determine the EV, and with the variance one can get a good picture of ruin and win goal probabilities from any length of play from it.
Improper house edge can underestimate the actual ev and the more trials, the greater the error becomes."
When calculating the EV over many bets, one needs to use the HE for bets resolved and not per roll unless one calculates how many rolls they will make first.
This is perfect.
Quote: AhighIn case you ever get to the point that you actually listen to what I am saying, realize that when you create your own bet that says you can hit either one of the four or ten before a seven, that new bet has a fewer average number of rolls.
The edge per roll is the same, OBVIOUSLY.
But when the average number of rolls diminish, the edge per event also diminishes.
I can't help but find humor in how long this is taking WITH ME LEADING YOU AND POINTING TO FAQ's.
But please, continue to pretend that I am wrong if it helps you get through the emotional parts of being wrong.
Edge per roll is 0.33333 for both bets.
Average number of rolls is 4 for just one numbers.
Average number of rolls is three when either number causes the bet to be over.
.333333 * 3 = 1.00
.333333 * 4 = 1.33
OMFG numbers!
Can you help walk me through this? Where does the .333 edge per roll number come from ? You can't make it too simple for me, so don't assume I understand anything other than the basics, ie 36 combinations, 3 ways to make a 4, etc.
Your method is just fine.Quote: RaleighCrapsI am assuming that the formula that I took from another posting is, in fact, THE correct formula.
(expected win - expected loss)/ Action = HE
From there, I used logic to determine what I felt the correct numbers should be for each variable.
Are there other acceptable ways to calculate HE?, OR
more importantly, is there a better way to calculate Expected Outcome?, since that is really all I care about.
What is the Expected Outcome if I Buy the 4 and 10 for $50 each, and take them both down on a Win of either number?
----------------------------------------------------------------------------------------------------------------------
same formula, different numbers......
$5000 Buy 4, pays $10,000 - $250 Vig = $9750
(.25*9750 + .25*9750 - .5*10000) / (.25*5000 + .25*5000 + .5*10000) = 1.667% HE
Your first formula was:
(.5*49- .5*50)/(.25*$25+.25*$25+.50*$50) = 1.33%
(.5*49- .5*50) the .5*49, the $49 is the actual win
(.25*9750 + .25*9750 - .5*10000)
You also show $9750 as the actual win.
Is it not just $4750, just like the $49?
I must be missing something here
Quote: guido111Your method is just fine.Quote: RaleighCrapsI am assuming that the formula that I took from another posting is, in fact, THE correct formula.
(expected win - expected loss)/ Action = HE
From there, I used logic to determine what I felt the correct numbers should be for each variable.
Are there other acceptable ways to calculate HE?, OR
more importantly, is there a better way to calculate Expected Outcome?, since that is really all I care about.
What is the Expected Outcome if I Buy the 4 and 10 for $50 each, and take them both down on a Win of either number?
----------------------------------------------------------------------------------------------------------------------
same formula, different numbers......
$5000 Buy 4, pays $10,000 - $250 Vig = $9750
(.25*9750 + .25*9750 - .5*10000) / (.25*5000 + .25*5000 + .5*10000) = 1.667% HE
Your first formula was:
(.5*49- .5*50)/(.25*$25+.25*$25+.50*$50) = 1.33%
(.5*49- .5*50) the .5*49, the $49 is the actual win
(.25*9750 + .25*9750 - .5*10000)
You also show $9750 as the actual win.
Is it not just $4750, just like the $49?
I must be missing something here
The $49 win is on a $25 bet.
The bet is $5000, therefore the win is $9,750
Got it!Quote: RaleighCraps
The bet is $5000, therefore the win is $9,750
Thanks
I must have looked at the number 5,000 times and only saw 4750 (Baccarat math got in the way)
Got a good laugh from it
You are now a house edge calculation expert, well at least for the per bet resolved part.
The per roll just uses different probabilities
Quote: guido111
1000 bets * $bet * HE = EV
Most use the 1.33% for the HE in the EV formula.
That $bet should be the average $bet resolved or you will end up with a value your simulations do not agree with. In your thread example that would be 25/2 + 50/2
No more or no less
Now, you could posit a combined bet of "place 6 or 8", which works just like it sounds, and it pays 7-to-12. You could also compute its house edge as (7/12)*(10/16)-(6/16)= 1.04%. So that's lower than the edge on the individual place bets. Does it make sense that a combination of two bets has a lower edge than either component? Not really, and it turns out that the comparison isn't a fair one because of the resolution time. Unlike the place bets which resolve every 3.27 rolls, this combined bet resolves every 36/16 = 2.25 rolls: much faster. On the same 100 rolls/hour table, the $12 combination has an expected loss of, yes, $5.56/hour.
In other words, this is all just a reminder to compare apples to apples -- when looking at two different bets, if they resolve at different speeds, the only accurate comparison is by EV/roll. In this case, 1.52% / 3.27 = 1.04% / 2.25. Both lose at 0.463%/roll. The same is true for the issue of combined buy bets or really any other combination. Just don't look at those lower numbers like 1.04% and think you've beaten the pass line. The pass line loses at about 0.42%/roll.
Quote: MathExtremistthe expected loss for a combination of bets is just the sum of the expected loss of the individual bets
Correct. My brain is not working good enough this morning to do any number crunching, but it occurred to me that HE, again, is not the sole thing to look at, where EV can give a quick answer.
Quote: Ahighyour chance of winning money goes up AND the edge goes DOWN by making more bets.
On the pass line you can bet the free odds, the most dramatic way to make the HE go down and a case of verifying the second part of that statement; whereas the fact that the HE can change with combinations, but not the EV of each bet, strikes down the first part of the statement even using the most powerful attack on the HE available in Craps, the free odds.
For the $25 Buy bets in question ev = -.3333 for one betQuote: MathExtremistIt's worth stepping back and remembering that this is all a (mostly useless) argument about the denominator. There is no dispute, at least there shouldn't be, that the expected loss for a combination of bets is just the sum of the expected loss of the individual bets.
$25 * (1/75)
2 bets = -.6666
Combined house edge -.6666/50
Variance = 1216.888889 for one bet (SD=34.88393454)
2433.777778 2 bets variance
SD for both bets: 49.33
Who says sd for both is 34.883*2
Guido asks about 1000 bets.
EV = N*HE*average$bet = -333.33
what no takers
someone says the HE is 1% so the EV = -250?
SD = 1560.056979 (sqrt(1000)*49.33)
probability of being even or ahead about 0.415400171
Not impossible by any means.
There are even better bets to show a profit after 1k wagers and same action.
=(0.5*49+0.5*-50)/(0.5*25+0.5*50)>>>>This one matches simulation results
=(0.5*49-0.5*50)/(50)>>>>This one matches no simulation results
In the end, no one cares
all that is left is the claim of one DI
Quote: 7crapsIn the end, no one cares
all that is left is the claim of one DI
It's not true that nobody cares. That only that would be true would be if:
A) I was not somebody
B) It didn't matter how you attacked me about this issue when I initially brought it up
The reason why I care is because, and as I said, people like yourself can and have been known to be really rude and arrogant to other people about details like this.
It has taken a really long time to get to the bottom of this issue, and I'm pretty sure that as I type, many people are still going to believe that I have been wrong in spite of the facts. I was banned for over two weeks as a result of how you decided to put me on an ignore list and post some pornographic image along with another image with my name on it that had an image URL that I inferred from that you thought that I was an "ass."
That was my "welcome wagon" to the Wizard of Odds. Your post there.
That's why I care. So the way I interpret "nobody cares" is that you don't care about me. And that much I can agree with.
You and guido also don't care to apologize about how you guys have tried so hard to discredit me on every single issue that you could possible hope to achieve to discredit me on.
Many people on here have tried very hard to associate me with others who charge money for classes.
And other such similar small-minded interpretations of what I am doing.
And so my point is this: I do care about how I am treated on this board.
And you and your comments such as this one that "nobody cares" is part of that treatment.
It's rude. It's inconsiderate. It's inappropriate. And it reflects poorly on you and others that blindly try to group me in with other people thinking about similar things to what I am thinking about.
And when these issues are resolved, the result is that "oh who cares" it really demonstrates that you are actually one of the types of persons who would like to just write off your losses and concentrate on your wins. Not me with the video camera. Because you'd like to think it doesn't matter when you lose.
You lost. That's why you wished that people like me didn't care. I am somebody, and I do care. So you're wrong about your assertion that "in the end, no one cares" as well.
If everybody could simply agree to only talk about edge per roll we wouldn't even be having these conversations.
The edge per roll for a $25 buy on the 4 with vig on the win is 1.33/4 = 0.33%
Is there anybody who disagrees with me on this point? Because if you can agree on that, as I think someone else pointed out, all of these other points of confusion are just that: confusion.
I've never been confused about the edge per roll being 0.33% for a $25 buy on the four. And I've never been confused about the 50/50 chance of hitting a four or a ten gets reduced to a 3 roll bet on average instead of a 4 roll bet on average.
But apparently some other people apply formulas that were designed for events, and they want to count two events when it's only one event and they get confused about how to set up the problem, and crank away on inputs that are wrong, and say they are right because they got a right answer with the wrong inputs.
A bet that has a house edge of 0.33% per roll and lasts for 3 rolls has a house edge of 1.00%.
Anybody who thinks that I'm wrong about that, just please explain it.
But the fact is and has been that I'm not wrong, I was never wrong, and if someone else was right about something else, they certainly weren't right about me being wrong, and that's what started this whole argument. Someone coming and taking a big poop on how I do math saying it was incorrect.
And now that we've gone through this over and over, nobody, and I mean nobody, is man enough to say "you know what Ahigh .. I just didn't understand you, but you were in fact right and I apologize for implying that you weren't as smart as me."
If there's anybody who doesn't know who might be in a position to say that, their name is 7 craps.
Quote: odiousgambitCorrect. My brain is not working good enough this morning to do any number crunching, but it occurred to me that HE, again, is not the sole thing to look at, where EV can give a quick answer.
On the pass line you can bet the free odds, the most dramatic way to make the HE go down and a case of verifying the second part of that statement; whereas the fact that the HE can change with combinations, but not the EV of each bet, strikes down the first part of the statement even using the most powerful attack on the HE available in Craps, the free odds.
The EV of each bet doesn't change.
The part that you and so many other people are having trouble with is that by taking both bets down on a win, and agreeing to operate that way, you no longer have the same two old bets, you a single new bet.
Got it.Quote: MathExtremistIt's worth stepping back and remembering that this is all a (mostly useless) argument about the denominator. There is no dispute, at least there shouldn't be, that the expected loss for a combination of bets is just the sum of the expected loss of the individual bets. If my expected loss rate on a given table for a buy 4 bet is, say, $10/hour, then if I make a buy 10 bet for the same amount, my loss rate doubles to $20/hour.
There now appears that 4 different people say the overall HE for the 2 Buy bets of $25 is 1.33%
One says .33%,1% and 1.33% but looks to still say 1.0% is also correct.
Simulations for the $25 and $5000 bets show 1.33% and 1.67% house edge values.
This makes calculating the total EV for 1000 such bets very easy.
again from
https://wizardofvegas.com/forum/gambling/betting-systems/11356-what-are-the-odds/3/#post183638
Quote: Ahigh
250/20000 = 1.25% edge for the event. Better than all on the pass. Better edge than all on the don't.
...
please correct me on anything I'm doing wrong here.
Quote: MathExtremistYour denominator is incorrectly inflated. You shouldn't include the action for the unresolved bets.
50% of the time you win 9750 on action of 5000, and 50% of the time you lose 10000 on action of 10000. EV = (0.5*9750 - 0.5*10000) / (0.5*5000 + 0.5*10000) = -125/7500 = -1.667%.
https://wizardofvegas.com/forum/gambling/craps/11405-craps-multiple-wagers-house-edge/3/#post184344
Quote: RaleighCraps
same formula, different numbers......
$5000 Buy 4, pays $10,000 - $250 Vig = $9750
(.25*9750 + .25*9750 - .5*10000) / (.25*5000 + .25*5000 + .5*10000) = 1.667% HE
Even with the $5000 Buy bets, not all agree that the correct answer is 1.67% for the house edge.
My sims still show the 1.67% and I can not get it down to 1,25%
So, the only conclusion to this point, IMO since not all can agree on how to determine the proper HE values for multiple craps bets that can accurately determine the EV for X number of such bets,
everyone can not be all correct.
MathExtremist has a great track record of very accurate math posts and lets one work out the problem so one can learn by doing, not just reading, and I am more inclined to stay on his side.
He has even shown 2 different ways to do the math.
I do agree that this is important.
RCraps also wants to know the right way to calculate these type of multiple wagers.
Some do not care. That is fine. They will not be invited to the Holiday party.
https://wizardofvegas.com/forum/gambling/craps/11405-craps-multiple-wagers-house-edge/2/#post184319
Quote: RaleighCraps...and I am genuinely asking you to tell me where my logic has gone wrong.
I am serious here.
I truly want to learn this stuff....
show me where my logic has gone wrong, so that I may continue to learn. Thanks
This is good stuff.
I think RaleighCraps deserves an answer to his post.
For us that do care and want our calculations to match our simulations so we can sleep better at night,
let us see who else wants to try.
Ahigh " You and guido also don't care to apologize about how you guys have tried so hard to discredit me on every single issue that you could possible hope to achieve to discredit me on. "
I feel this calls for a 3 day suspension for Ahigh contradicting Ahigh !
Everything we are arguing about, again, is that I am talking about a single bet constructed from two bets. My bet lasts a shorter number of rolls. The edge per roll is agreed upon at 0.33%. The number of rolls is 3 instead of 4.
It's simple.
The simulator should verify that the edge per roll is 0.33% and the number of times you get paid plus the number of sevens is the number of events.
You get paid exactly as often as you lose. That's your 50/50 wager.
You have one or the other thing happen 12 out of 36 times, or 1 in 3 rolls.
Nice that even the math folks have those easy to understand words, "expect" / "expectation" / "negative"
The WizardQuote: AhighAbout the "you shouldn't include the action for the unresolved bet" .. the Wizard explains this in the FAQ. You can include the action if you agree at the beginning to take the bet down on the win .. when you take it down, that resolves it.
Everything we are arguing about, again, is that I am talking about a single bet constructed from two bets. My bet lasts a shorter number of rolls. The edge per roll is agreed upon at 0.33%. The number of rolls is 3 instead of 4.
It's simple.
The simulator should verify that the edge per roll is 0.33% and the number of times you get paid plus the number of sevens is the number of events.
"However in this case the player is only keeping the place bets up for one roll."
And this is exactly the same for Buying the 4 and the 10? One roll and they come down?
He was also just showing the HE to 1.136% and that is only a per roll value.
That is way easier to use to determine total EV when part of the IC resolves every roll but not every bet of the IC does so.
The per bet house edge is 2.486%
The iron cross 2 house edges were talked about here
https://wizardofvegas.com/forum/questions-and-answers/math/3573-house-vig-on-anything-but-7-rdutch/#post40530
https://wizardofvegas.com/forum/questions-and-answers/math/3573-house-vig-on-anything-but-7-rdutch/2/#post40763
Back to this threads topic.
Multiple Buy bets house edge.
My simulations still verifies a house edge of 1.33% over 10,000,000 resolved bets and not the 1% you claim but the 1.33% that you showed in one of your posts.
The mean (per 1,000 bets) is very steady at -500
If the true HE was 1% per bet resolved, as you only claim,
why is the EV not -375?
We are all
still not correct.
You have now added that the Wizard says it is OK to determine the HE at 1%,
so then are we to also say the Wizard believes it is also 1%?
I do not think he would like someone stating something for him, but maybe he does based from his earlier writings that are not the same as this threads Craps Buy bet example.
Any other's simulations results out there?
Per bet resolved or per roll.
New ideas welcomed.
Wonder why M146 has not attempted the math here?
If you work out the mathematics of dice control, you will find that it only takes one controlled throw every 43 rolls of the dice to eliminate the house edge for the 6 and 8 place bets.
Many want AHigh's hardway talent that Frank and Co do not possess.
Guess we just have to be at the casino when he is there and wing it.
Should be easy, he says he plays a lot.
He does not want a Craps Event of the Century to be filmed in an actual casino.
OK
But he still has the needed talent to overcome the 9to11% HE for the hardway bets,
at least that is my current understanding from one of RCraps posts.
All I want is the money.
More money.
easy money.
Quote: guido111
Wonder why M146 has not attempted the math here?
Math?
I'm having trouble understanding what all this fuss is about.
Here's the Math:
(1.33 + 1.33) /2 = 1.33
Done.
Quote: Mission146Math?
I'm having trouble understanding what all this fuss is about.
Here's the Math:
(1.33 + 1.33) /2 = 1.33
Done.
I am Buzz Paff and I approve this message .
Thanks.Quote: Mission146Math?
I'm having trouble understanding what all this fuss is about.
Here's the Math:
(1.33 + 1.33) /2 = 1.33
Done.
OK. That is your answer for question #2.
That seems to make about 5 that agree on that value.
Problem is not all agree with that 1.33% HE value.
#2) But [sic] the 4 AND 10 for $25 each and take $99 and down on the first hit, the edge is 1/100 or 1.00%
The 1% still holds water by AHigh because from a per roll times the expected number of rolls he arrives at 1%.
I have tried different methods of making the bet, taking down after a win, waiting for a 7, replacing the bets, no waiting for a 7 and my sims still show 1.33%.
AHigh also says the Wizards method of calculating house edge, as was done for the Iron Cross bet,
when bets do not resolve on every roll proves his 1% HE value.
Unless this thread is deleted, it is still live for more interesting conclusions to the mystery of the 50 series for two $25 Buy bets.
Quote: Mission146Math?
I'm having trouble understanding what all this fuss is about.
Here's the Math:
(1.33 + 1.33) /2 = 1.33
Done.
The reason you're wrong is that this is the edge for both bets resolved. However, if one wins, you're taking them down with the other unresolved, and this decreases the edge. Treat them separately -
Win (+49) on (4/10)
Take down on (10/4)
Loss (-25) on 7
3 winning combinations, +147, 6 losing combinations, -150, 12 total "playing" combinations. EV by the time it's down is -3/12 = 25¢, on a $25 bet, for a 1% house edge. (1 + 1)/2 = 1.
Of course, if you keep this up for a session, you'll find the edge per roll was exactly the same.
The loss is $50.Quote: 24BingoWin (+49) on (4/10)
Take down on (10/4)
Loss (-25) on 7
3 winning combinations, +147, 6 losing combinations, -150, 12 total "playing" combinations. EV by the time it's down is -3/12 = 25¢, on a $25 bet, for a 1% house edge. (1 + 1)/2 = 1.
There are Two $25 Buy bets there.
Mission 146 is adding house edge, that is the wrong method to use as stated by ME.
We add ev and variance not house edge and standard deviation.
Try one $25 Buy4 and one $100 Buy10 and see what you get adding edges.
The solution to this madness is AHigh answering RaleighCraps post on how he (AHigh)
calculated the per roll edge.
Quote: odiousgambitMr. Hightower, do you understand that you cannot change the negative expectation of each bet? Thus, if you bet more, you may be able to lower the house edge vis a vis your total bets, but it is still true that the more you bet on various combinations of bets with house edge, the more you are expected to lose.
Nice that even the math folks have those easy to understand words, "expect" / "expectation" / "negative"
Absolutely agree that you can expect to lose more money the more you put up there for every non-free bet on the table.
The events just get shorter. That's what makes the number go down: they resolve more quickly.
Again, it all hinges around how you define your event.
Edge per roll never goes below 0.33 for any non-zero edge bet on the craps table that I know of, and the buy on the four and ten for a quarter beats every other bet out there that I know of (unless you can pay $1 vig for more than a $25 bet).
As far as the simulator, you would have to define the bet as winning on a 4 or a 10 and losing on a seven to get the 1.00% edge.
If you show one bet resolving on a 4 and another bet resolving on the 10 and both bets resolving on the seven, you're going to get the 1.33% number.
Your simulator needs to have a definition of a new bet that is called "4or10" in order to simulate the answer to come out right.
I say the bet is possible whether they mark it or not. It is a bet that you can make. But for it to be that bet, you have to define the bet as being done when either one of the 4 or 10 hits.
We'll have to leave it at that.
But to help people that want to understand imagine if you had no bet on the felt for doing a place bet on the 9.
But rather instead you could place the 5-4, and you could place the 6-3. Each would be a $2.50 bet that would pay $7.00. They would both lose on a seven.
So the 5-4 edge would be $0.50 / $2.50 + $7.50 = 5% house edge.
And the 6-3 edge would also be $0.50 / $2.50 + $7.50 = 5% house edge
by betting them both and agreeing the either one of them hitting would end the event, you could have what we currently know as placing the 9.
And that has an edge of $0.50 / ( $5.00 + $7.50 ) = 4% house edge.
If you can understand this, maybe it will help understand the other.
The only difference is that this bet is marked on the felt, and you can't place just half of the 5 or half of the 9. But you BUY half of the "4or10" bet as I define it.
Okay?
Quote: AhighEdge per roll never goes below 0.33 for any non-zero edge bet on the craps table that I know of, and the buy on the four and ten for a quarter beats every other bet out there that I know of (unless you can pay $1 vig for more than a $25 bet).
The Santa Ana Star in New Mexico offers free (no-vig) buy bets on the 4 and 10, and also offers a zero-edge field bet that pays triple on both 2 *and* 12.
http://www.santaanastar.com/gaming/table-games/craps.
Will the Silverton let you go higher than $25 on the 4 or 10 for $1 vig? I've heard some casinos will go as high as $39 because they take the floor of the vig rather than rounding to the nearest dollar. I've also heard that some casinos will treat each bet buy separately for vig calculations vs. others lumping them together. In other words, you'd pay $4 vig on $50 each buy 4 and 10 vs. $5 vig, which is what you'd pay if you put all $100 on one number. But that varies from property to property and is almost always something you have to ask.
